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Question-150062

Question Number 150062 by 0731619 last updated on 09/Aug/21 Answered by Ar Brandon last updated on 09/Aug/21 $${li}_{\mathrm{2}} \left({x}\right)\underset{{function}} {\overset{{dilogarithm}} {=}}\underset{{n}=\mathrm{1}} {\overset{\infty} {\sum}}\frac{{x}^{{n}} }{{n}^{\mathrm{2}} }=−\int_{\mathrm{0}}…

Question-84496

Question Number 84496 by bshahid010@gmail.com last updated on 13/Mar/20 Commented by mathmax by abdo last updated on 13/Mar/20 $${I}\:=\int\:\:\frac{{arcsinx}}{{x}^{\mathrm{2}} }{dx}\:\:{by}\:{parts}\:\:{u}^{'} =\frac{\mathrm{1}}{{x}^{\mathrm{2}} }\:{and}\:{v}={arcsinx}\:\Rightarrow \\ $$$${I}\:=−\frac{{arsinx}}{{x}}\:−\int−\frac{\mathrm{1}}{{x}}\:×\frac{\mathrm{1}}{\:\sqrt{\mathrm{1}−{x}^{\mathrm{2}} }}{dx}…

1-dx-1-tanx-2-tanx-sinx-cosx-dx-

Question Number 149996 by tabata last updated on 08/Aug/21 $$\left(\mathrm{1}\right)\:\int\:\:\frac{{dx}}{\mathrm{1}+{tanx}} \\ $$$$ \\ $$$$\left(\mathrm{2}\right)\int\:\:\frac{\sqrt{{tanx}}}{{sinx}\:{cosx}}{dx} \\ $$ Answered by mindispower last updated on 08/Aug/21 $$=\int\frac{\mathrm{1}}{{cos}^{\mathrm{2}} \left({x}\right)}.\frac{\sqrt{{tg}\left({x}\right)}}{{tg}\left({x}\right)}{dx}=\int\frac{\mathrm{1}}{{cos}^{\mathrm{2}}…

By-subs-u-2-4-x-evaluate-4-x-x-dx-

Question Number 149989 by ZiYangLee last updated on 08/Aug/21 $$\mathrm{By}\:\mathrm{subs}\:{u}^{\mathrm{2}} =\mathrm{4}+{x},\:\mathrm{evaluate}\:\int\:\frac{\sqrt{\mathrm{4}+{x}}}{{x}}\:{dx} \\ $$ Answered by puissant last updated on 08/Aug/21 $$\int\frac{\sqrt{\mathrm{4}+{x}}}{{x}}{dx}={Q} \\ $$$${u}=\sqrt{\mathrm{4}+{x}}\:\rightarrow\:{u}^{\mathrm{2}} =\mathrm{4}+{x}\:\rightarrow\:{x}={u}^{\mathrm{2}} −\mathrm{4}…

Question-149946

Question Number 149946 by DELETED last updated on 08/Aug/21 Answered by DELETED last updated on 08/Aug/21 $$\boldsymbol{\mathrm{Jawaban}}\:\boldsymbol{\mathrm{no}}\:\mathrm{2} \\ $$$$\mathrm{Gunakan}\:\mathrm{Hk}\:\mathrm{Kirchoff}\:\mathrm{I} \\ $$$$\Sigma\mathrm{I}_{\mathrm{masuk}} =\Sigma\mathrm{I}_{\mathrm{keluar}} \\ $$$$\mathrm{I}_{\mathrm{1}} +\mathrm{I}_{\mathrm{2}}…